Do the $p$-adic numbers $K$ have the property that for $f \in x + (x^2)(K[x])$ then $f : K \rightarrow K$ is injective on some neighborhood of $0$?

Question

For a topological ring $K$, let us say $K$ has property $P$ if for each $f \in x + (x^2)(K[x]) \subset K[x]$, such $f$ is injective on some neighborhood of $0 \in K$ when viewed as a function $f:K\rightarrow K$.

As I understand, if a topological ring has property $P$ then any topological subring of it does as well. Furthermore, the field of complex numbers (as well as the reals, both with the usual topologies) has property $P$.

My question is whether the field of $p$-adic numbers (with its usual absolute value, making it a topological field) has property $P$? How would this be shown? (And if yes, does a similar result also hold for more general Hausdorff topological fields?)

Answer 1

Definition: Let $(K, \lvert \cdot \rvert)$ be a field complete w.r.t. a nonarchimedean absolute value. If $X \neq \emptyset$ is a subset of $K$ without isolated points, $f:X \rightarrow K$ is called strictly differentiable at $a \in X$ if $$\lim_{(x,y)\to (a,a)}\dfrac{f(y)-f(x)}{y-x}$$ exists. As the name suggests, this implies (but is stronger than) that $f$ is differentiable at $a$, and the limit in question is $f'(a)$.

Fact 1: Polynomial functions are strictly differentiable everywhere. In fact, analytic functions are strictly differentiable inside their domain of convergence.

Fact 2: If $f$ is strictly differentiable at some $a \in X$ and $f'(a) \neq 0$, then there is a neighbourhood $U$ of $a$ in $X$ such that $$\lvert f(y)-f(x)\rvert = \lvert f'(a) \rvert \lvert y-x \rvert $$ for all $x,y \in U$; in particular, $f$ is injective on $U$.

To show Fact 2, one needs the strict triangle inequality. But even without this, one gets injectivity on some small neighbourhood of $a$ from strict differentiability with nonzero derivative: Because if every neighbourhood of $a$ contained two different points $x\neq y$ with $f(x)=f(y)$, then if the limit in the definition exists at all, it must be $0$.

The whole point of strict differentiability is to exclude things like

Example: The following function $f:\mathbb Z_p \rightarrow \mathbb Q_p$ has derivative $f'(x)=1$ everywhere, but is not injective in any neighbourhood of $0$. (In particular, it is not strictly differentiable at $0$.)

$$f(x) = \begin{cases} x-p^{2n} \quad \text{ if } \lvert x - p^n \rvert \le p^{2n+1}\\ x \qquad \text{ else} \end{cases}$$

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All this is beautifully explained in W. H. Schikhof, Ultrametric Calculus: An Introduction to p-Adic Analysis (Cambridge Studies in Advanced Mathematics. Cambridge: Cambridge University Press. https://doi.org/10.1017/CBO9780511623844). (One unfortunate thing about this source is that he tried to call strict differentiability $C^1$ which has not been followed in the literature since.) The above definition, facts 1 and 2, and the example, are Definition 27.1, Exercise 27.A, Proposition 27.3, and Example 26.6. in that book, respecively.